$\begin {array} {1 1} (A)\;tautology & \quad (B)\;fallacy \\ (C)\;cannot \: be\: determined & \quad (D)\;None\: of\: these \end {array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Ans : (A)

Assume that $ a \geq b \geq c$. We must have $b+c > a.$

Also note that $b+c \leq c+a \leq a+b$

$ \large\frac{1}{(b+c)} \geq \large\frac{ 1}{(c+a)} \geq \large\frac{1}{(a+b)}$

To show that $ \large\frac{1}{(b+c)},\large\frac{1}{(c+a)},\large\frac{1}{(a+b)}$ are sides of a triangle,

it is sufficient to show that

$ \large\frac{1}{(c+a)} + \large\frac{1}{(a+b)} > \large\frac{1}{(b+c)}$……………(i)

As $a \geq b \geq c$, we get

$2a \geq a+b\: and\: 2a \geq a+c$

$\large\frac{1}{2a} \leq \large\frac{1}{(a+b)}, \large\frac{1}{2a} \leq \large\frac{1}{(a+c)}$

$\large\frac{1}{(a+b)} + \large\frac{1}{(a+c)} \geq \large\frac{1}{2a} + \large\frac{1}{2a} = \large\frac{1}{a} > \large\frac{1}{(b+c)}$ [From Eq. (i)]

So, it represents a triangle.

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...